Question

use a right triangle to write the following expression as an algebraic expression. assume that x is positive and in the domain of the given inverse trigonometric function. tan(cos^(-1)9x) which of the following triangles is correct to write the given expression as an algebraic expression? a. triangle with sides labeled b. triangle with sides labeled c. triangle with sides labeled d. triangle with sides labeled tan(cos^(-1)9x)= (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators)

Explanation:

Step1: Recall inverse - cosine definition

Let $\theta=\cos^{-1}(9x)$. Then, by the definition of the inverse - cosine function, $\cos\theta = 9x=\frac{9x}{1}$, where in a right - triangle, if the adjacent side to the angle $\theta$ is $9x$ and the hypotenuse is $1$.

Step2: Find the opposite side using Pythagorean theorem

Let the opposite side be $y$. By the Pythagorean theorem $a^{2}+b^{2}=c^{2}$, where $a = 9x$, $c = 1$. So $y=\sqrt{1-(9x)^{2}}=\sqrt{1 - 81x^{2}}$.

Step3: Recall the tangent function definition

We know that $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. Since $\theta=\cos^{-1}(9x)$ and the opposite side is $\sqrt{1 - 81x^{2}}$ and the adjacent side is $9x$, then $\tan(\cos^{-1}(9x))=\frac{\sqrt{1 - 81x^{2}}}{9x}$.

For the correct right - triangle, if $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} = 9x=\frac{9x}{1}$, the adjacent side to the angle $\theta$ is $9x$ and the hypotenuse is $1$. So the correct triangle is the one where the adjacent side to the angle $\theta$ is $9x$ and the hypotenuse is $1$. Looking at the options, we need to find the triangle with adjacent side $9x$ and hypotenuse $1$.

Answer:

If we assume the options are labeled in order from top - to - bottom as A, B, C, D, the correct triangle is the one where the adjacent side to the angle $\theta$ is $9x$ and the hypotenuse is $1$. And $\tan(\cos^{-1}(9x))=\frac{\sqrt{1 - 81x^{2}}}{9x}$